\(\int \frac {x (1+x^2)^3}{(2+2 x^2+x^4)^2} \, dx\) [171]

Optimal result

Integrand size = 21, antiderivative size = 32 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {1}{4 \left (2+2 x^2+x^4\right )}+\frac {1}{4} \log \left (2+2 x^2+x^4\right ) \]

[Out]

1/4/(x^4+2*x^2+2)+1/4*ln(x^4+2*x^2+2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1261, 700, 642} \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {1}{4} \log \left (x^4+2 x^2+2\right )-\frac {\left (x^2+1\right )^2}{4 \left (x^4+2 x^2+2\right )} \]

[In]

Int[(x*(1 + x^2)^3)/(2 + 2*x^2 + x^4)^2,x]

[Out]

-1/4*(1 + x^2)^2/(2 + 2*x^2 + x^4) + Log[2 + 2*x^2 + x^4]/4

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 700

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*(d + e*x)^(m - 1)*(
(a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] - Dist[d*e*((m - 1)/(b*(p + 1))), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2
*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(1+x)^3}{\left (2+2 x+x^2\right )^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (1+x^2\right )^2}{4 \left (2+2 x^2+x^4\right )}+\frac {1}{2} \text {Subst}\left (\int \frac {1+x}{2+2 x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (1+x^2\right )^2}{4 \left (2+2 x^2+x^4\right )}+\frac {1}{4} \log \left (2+2 x^2+x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {1}{4} \left (\frac {1}{1+\left (1+x^2\right )^2}+\log \left (1+\left (1+x^2\right )^2\right )\right ) \]

[In]

Integrate[(x*(1 + x^2)^3)/(2 + 2*x^2 + x^4)^2,x]

[Out]

((1 + (1 + x^2)^2)^(-1) + Log[1 + (1 + x^2)^2])/4

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {1}{4 x^{4}+8 x^{2}+8}+\frac {\ln \left (x^{4}+2 x^{2}+2\right )}{4}\)	\(29\)
norman	\(\frac {1}{4 x^{4}+8 x^{2}+8}+\frac {\ln \left (x^{4}+2 x^{2}+2\right )}{4}\)	\(29\)
risch	\(\frac {1}{4 x^{4}+8 x^{2}+8}+\frac {\ln \left (x^{4}+2 x^{2}+2\right )}{4}\)	\(29\)
parallelrisch	\(\frac {\ln \left (x^{4}+2 x^{2}+2\right ) x^{4}+1+2 \ln \left (x^{4}+2 x^{2}+2\right ) x^{2}+2 \ln \left (x^{4}+2 x^{2}+2\right )}{4 x^{4}+8 x^{2}+8}\)	\(61\)

[In]

int(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x,method=_RETURNVERBOSE)

[Out]

1/4/(x^4+2*x^2+2)+1/4*ln(x^4+2*x^2+2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {{\left (x^{4} + 2 \, x^{2} + 2\right )} \log \left (x^{4} + 2 \, x^{2} + 2\right ) + 1}{4 \, {\left (x^{4} + 2 \, x^{2} + 2\right )}} \]

[In]

integrate(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x, algorithm="fricas")

[Out]

1/4*((x^4 + 2*x^2 + 2)*log(x^4 + 2*x^2 + 2) + 1)/(x^4 + 2*x^2 + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {\log {\left (x^{4} + 2 x^{2} + 2 \right )}}{4} + \frac {1}{4 x^{4} + 8 x^{2} + 8} \]

[In]

integrate(x*(x**2+1)**3/(x**4+2*x**2+2)**2,x)

[Out]

log(x**4 + 2*x**2 + 2)/4 + 1/(4*x**4 + 8*x**2 + 8)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {1}{4 \, {\left (x^{4} + 2 \, x^{2} + 2\right )}} + \frac {1}{4} \, \log \left (x^{4} + 2 \, x^{2} + 2\right ) \]

[In]

integrate(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x, algorithm="maxima")

[Out]

1/4/(x^4 + 2*x^2 + 2) + 1/4*log(x^4 + 2*x^2 + 2)

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {1}{4 \, {\left (x^{4} + 2 \, x^{2} + 2\right )}} - \frac {1}{4} \, \log \left (\frac {1}{2 \, {\left (x^{4} + 2 \, x^{2} + 2\right )}}\right ) \]

[In]

integrate(x*(x^2+1)^3/(x^4+2*x^2+2)^2,x, algorithm="giac")

[Out]

1/4/(x^4 + 2*x^2 + 2) - 1/4*log(1/2/(x^4 + 2*x^2 + 2))

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (1+x^2\right )^3}{\left (2+2 x^2+x^4\right )^2} \, dx=\frac {\ln \left (x^4+2\,x^2+2\right )}{4}+\frac {1}{4\,\left (x^4+2\,x^2+2\right )} \]

[In]

int((x*(x^2 + 1)^3)/(2*x^2 + x^4 + 2)^2,x)

[Out]

log(2*x^2 + x^4 + 2)/4 + 1/(4*(2*x^2 + x^4 + 2))